[seqfan] Re: Number of ways to write n^2 as the sum of n odd numbers
Richard Mathar
mathar at strw.leidenuniv.nl
Fri Sep 25 20:44:12 CEST 2009
dr> Date: Fri, 25 Sep 2009 12:11:17 -0500
dr> From: David Radcliffe <dradcliffe at gmail.com>
dr> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
dr> Subject: [seqfan] Number of ways to write n^2 as the sum of n odd numbers
dr>
dr> Greetings Sequence Fans,
dr>
dr> Everybody knows that the nth square is equal to the sum of the first n
dr> odd numbers.
dr> It might be interesting to calculate the number of ways to write n^2
dr> as the sum of n
dr> odd numbers, disregarding order.
dr>
dr> For example, 9 can be written as a sum of three odd numbers in 3 ways: 1+1+7,
dr> 1+3+5, and 3+3+3.
dr>...
An even more fundamental question is whether the table of the number
of partitions of n into k odd parts is in the OEIS. There are many ways
to write such a table. The row sums are A00009, and the table would start
with rows n>=1 and columns 1<=k<=n as
1
0 1
0 0 1
0 1 0 1
0 0 1 0 1
0 1 0 1 0 1
0 0 2 0 1 0 1
0 1 0 2 0 1 0 1
0 0 2 0 2 0 1 0 1
0 1 0 3 0 2 0 1 0 1
0 0 3 0 3 0 2 0 1 0 1
0 1 0 4 0 3 0 2 0 1 0 1
0 0 3 0 5 0 3 0 2 0 1 0 1
0 1 0 5 0 5 0 3 0 2 0 1 0 1
0 0 4 0 6 0 5 0 3 0 2 0 1 0 1
0 1 0 7 0 7 0 5 0 3 0 2 0 1 0 1
0 0 4 0 9 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 8 0 10 0 7 0 5 0 3 0 2 0 1 0 1
0 0 5 0 11 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 10 0 13 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 5 0 15 0 14 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 12 0 18 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 6 0 18 0 20 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 14 0 23 0 21 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 6 0 23 0 26 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 16 0 30 0 28 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 7 0 27 0 35 0 29 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 19 0 37 0 38 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 7 0 34 0 44 0 40 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 21 0 47 0 49 0 41 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 8 0 39 0 58 0 52 0 42 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 24 0 57 0 65 0 54 0 42 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 8 0 47 0 71 0 70 0 55 0 42 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 27 0 70 0 82 0 73 0 56 0 42 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 9 0 54 0 90 0 89 0 75 0 56 0 42 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 1 0 30 0 84 0 105 0 94 0 76 0 56 0 42 0 30 0 22 0 15 0 11 0 7 0 5 0 3 0 2 0 1 0 1
0 0 9 0 64 0 110 0 116 0 97 0 77 0 56 0 42 0 30 0 22 0 15 0 11 0 7 0 5 0
To eliminate the regular zeros, one would write it as triangle for the
odd n=1,3,5,... and odd k=1,3,5,...n (row sums A035294):
1
0 1
0 1 1
0 2 1 1
0 2 2 1 1
0 3 3 2 1 1
0 3 5 3 2 1 1
0 4 6 5 3 2 1 1
0 4 9 7 5 3 2 1 1
0 5 11 11 7 5 3 2 1 1
0 5 15 14 11 7 5 3 2 1 1
0 6 18 20 15 11 7 5 3 2 1 1
0 6 23 26 22 15 11 7 5 3 2 1 1
0 7 27 35 29 22 15 11 7 5 3 2 1 1
0 7 34 44 40 30 22 15 11 7 5 3 2 1 1
0 8 39 58 52 42 30 22 15 11 7 5 3 2 1 1
0 8 47 71 70 55 42 30 22 15 11 7 5 3 2 1 1
0 9 54 90 89 75 56 42 30 22 15 11 7 5 3 2 1 1
0 9 64 110 116 97 77 56 42 30 22 15 11 7 5 3 2 1 1
0 10 72 136 146 128 100 77 56 42 30 22 15 11 7 5 3 2 1 1
.. David's version is reading this version along a parabolic line n=k^2....
and another embedded triangle for the even n=2,4,6,8,.. and even k=2,4,6,..,n,
row sums A078408:
1
1 1
1 1 1
1 2 1 1
1 3 2 1 1
1 4 3 2 1 1
1 5 5 3 2 1 1
1 7 7 5 3 2 1 1
1 8 10 7 5 3 2 1 1
1 10 13 11 7 5 3 2 1 1
1 12 18 15 11 7 5 3 2 1 1
1 14 23 21 15 11 7 5 3 2 1 1
1 16 30 28 22 15 11 7 5 3 2 1 1
1 19 37 38 30 22 15 11 7 5 3 2 1 1
1 21 47 49 41 30 22 15 11 7 5 3 2 1 1
1 24 57 65 54 42 30 22 15 11 7 5 3 2 1 1
1 27 70 82 73 56 42 30 22 15 11 7 5 3 2 1 1
1 30 84 105 94 76 56 42 30 22 15 11 7 5 3 2 1 1
1 33 101 131 123 99 77 56 42 30 22 15 11 7 5 3 2 1 1
1 37 119 164 157 131 101 77 56 42 30 22 15 11 7 5 3 2 1 1
It seems in both sub-triangles, reading columns downwards "converges"
to A000041.
Richard J. Mathar
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